If  ${\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3=-{\mathrm{t}}_1{\mathrm{t}}_2{\mathrm{t}}_3$ then the orthocentre of the triangle formed by the points  $\mathrm{A}\left({\mathrm{at}}_1,\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)\right)\mathrm{B}\left({\mathrm{at}}_2{\mathrm{t}}_3,\mathrm{a}\left({\mathrm{t}}_2+{\mathrm{t}}_3\right)\right),\mathrm{C}\left({\mathrm{at}}_3{\mathrm{t}}_1,\mathrm{a}\left({\mathrm{t}}_3+{\mathrm{t}}_1\right)\right)$ , lies on

detailed solution

Correct option is A

Equation of altitude from the point A at1t2,at1+t2 is  y−at1+t2=−t3x−at1t2Re write this equation as     t3x+y=at1+t2+at1t2t3   …… (1) Equation of altitude from the point Bat2t3,at2+t3 is     t1x+y=at2+t3+at1t2t3   …… (2) The ortho center is the point of intersection of the above two lines Hence, subtract equation (1) from (2)  xt3−t1=at1−t3⇒x=−aSubstitute x=-a  in the equation t3x+y=at1+t2+at1t2t3t3(−a)+y=at1+t2+at1t2t3y=at1+t2+t3+at1t2t3=0Therefore, the ortho center is −a,0  it lies on x- axis.